Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int const N = 111111;
int const INF = 1e9 + 7;
int n, m, q;
vector<int> g[N];
int tin[N], tout[N], par[N][18];
int LOG;
int timer;
vector<int> people[N], path[N][18];
vector<int> ans;
vector<int> my_merge(vector<int> fi, vector<int> se) {
  sort(fi.begin(), fi.end());
  sort(se.begin(), se.end());
  vector<int> cur;
  cur.resize(fi.size() + se.size());
  merge(fi.begin(), fi.end(), se.begin(), se.end(), cur.begin());
  auto it = unique(cur.begin(), cur.end());
  cur.resize(distance(cur.begin(), it));
  cur.resize(min((int)cur.size(), 10));
  return cur;
}
void dfs(int v, int p = 1) {
  timer++;
  tin[v] = timer;
  par[v][0] = p;
  path[v][0] = my_merge(people[v], people[p]);
  for (int c = 1; c <= LOG; c++) {
    par[v][c] = par[par[v][c - 1]][c - 1];
    path[v][c] = my_merge(path[v][c - 1], path[par[v][c - 1]][c - 1]);
    assert(path[v][c].size() <= 10);
  }
  for (int c = 0; c < g[v].size(); c++) {
    int to = g[v][c];
    if (to == p) continue;
    dfs(to, v);
  }
  tout[v] = timer;
}
bool is_parrent(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int LCA(int a, int b) {
  if (is_parrent(a, b)) return a;
  if (is_parrent(b, a)) return b;
  for (int c = LOG; c >= 0; c--) {
    if (!is_parrent(par[a][c], b)) a = par[a][c];
  }
  return par[a][0];
}
void get(int a, int b) {
  if (a == b) {
    ans = my_merge(ans, people[a]);
    return;
  }
  for (int c = LOG; c >= 0; c--) {
    if (!is_parrent(par[a][c], b)) {
      ans = my_merge(ans, path[a][c]);
      a = par[a][c];
    }
  }
  ans = my_merge(ans, path[a][0]);
}
int main() {
  scanf("%d %d %d", &n, &m, &q);
  LOG = (int)log2(n);
  for (int c = 1; c < n; c++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int c = 1; c <= m; c++) {
    int x;
    scanf("%d", &x);
    people[x].push_back(c);
  }
  for (int c = 1; c <= n; c++) {
    if (people[c].size() > 10) {
      sort(people[c].begin(), people[c].end());
      people[c].resize(10);
    }
  }
  dfs(1);
  for (int c = 1; c <= q; c++) {
    int x, y, a;
    scanf("%d %d %d", &x, &y, &a);
    int lca = LCA(x, y);
    ans.clear();
    get(x, lca);
    get(y, lca);
    ans.resize(min((int)ans.size(), a));
    printf("%d ", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    puts("");
  }
  return 0;
}